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algol13
2 years ago
5

A High-speed internet provider A has a $100 setup fee and costs $65 per month. High-speed internet

Mathematics
2 answers:
Roman55 [17]2 years ago
4 0

Answer:

265

Step-by-step explanation:

i don't know I just added all of the answers up. 100+65=165. 70+30=100. 100+165=265

vodomira [7]2 years ago
4 0

Answer:

In 14 months, both providers will cost $1010

Step-by-step explanation:

Provider A costs 65 per month along with a 100 setup fee. You could write that as a slope-intercept equation, c = 65m+100
Provider B costs 70 per month along with a 70 setup fee. You could write that as a slope-intercept equation, c = 70m+30

You can put the two equations together, which would be 65m+100=70m+30. With simple algebra, you would get 5m=70. m would equal 14. So, after 14 months, the providers will cost the same. Now you can substitute 14 as m to find the price. c = 65(14)+100 = 1010. So, after 14 months, both providers will cost $1010.

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Answer:

D) 1/5 e 1/3

Step-by-step explanation:

You have the following quadratic equation:

15x^2-kx+1=0           (1)

In order to find the values of x that are solution to the equation (1), you first find the solution for k in the following equation:

2(k-8)+3(-k+1)=-4k+11\\\\2k-16-3k+3=-4k+11\\\\2k-3k+4k=11+16-3\\\\3k=24\\\\k=8

Next, you replace the previous value of k in the equation (1) and you use the quadratic formula to find the roots:

15x^2-8x+1=0\\\\x_{1,2}=\frac{-(-8)\pm \sqrt{(-8)^2-4(15)(1)}}{2(15)}\\\\x_{1,2}=\frac{8\pm 2}{30}\\\\x_1=\frac{1}{5}\\\\x_2=\frac{1}{3}

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